0
$\begingroup$

I have a doubt about photoelectric effect and the nature of light in general. From what I understood, in order to ionize a piece of some material, I need an electromagnetic wave with a frequence greater or equal than a certain frequence, meanwhile the intensity of the incident wave doesn’t matter. But does the intensity of the wave transfer some energy to the material (like the one calculated with the poynting vector?)? If I increase the intensity of the wave (i.e. its amplitude) will this energy be able to ionize the material at a certain point? So the final question might be: does an electromagnetic wave possess two energy? One associated with the frequency and one associated with the intensity?

$\endgroup$
  • $\begingroup$ The photoelectric effect demonstrates that electromagnetic waves are not classical but quantum mechanical phenomena. Quantum electrodynamics has been tested with experiments which have up to 13 digits of precision, or so, while semi-classical explanations of the photo-effect are struggling to fit the facts. Worse, they can't reproduce the precision tests, at all. Quantum field theory in general has been remarkably successful to explain the structure of matter and the spectrum of the physical vacuum up to the TeV range, while classical field theory can't even explain the stability of matter. $\endgroup$ – CuriousOne Oct 29 '15 at 9:31
  • $\begingroup$ It is true that it's possible to ionize with photons of lower energy than the photoelectric effect requires, but only at very high incident power levels. It's a multi-photon-absorption condition, and not easy to achieve. $\endgroup$ – Carl Witthoft Oct 29 '15 at 11:44
2
$\begingroup$

If you want to understand the photoelectric effect and energy of light you have to be very very careful to keep the terminology correct.

The energy of a electromagnetic wave (1 photon) is completely determined by its frequency.

The intensity of the electromagnetic wave(s) is how many photons per second are hitting on some area of surface. You can't combine photons to get a "bigger" photoelectric effect.

The gist of the photoelectric effect is this. The most loosely bound electrons in the material still have some specific binding energy. Until the photon has at least this much energy then it can't knock an electron free. But if it has more than that energy, then the electron gets the excess energy.

So if the binding energy is 4 eV:

  • A 3 eV photon doesn't knock out an electron
  • A 10 eV photon knocks out an electron with 6 eV of kinetic energy

The other "kind of energy transfer" (from the intensity of the photons) creates heat. So if a huge number of photons were absorbed by the material then it gets hot. If the material gets hot enough then it will emit electrons. (But this is not the photoelectric effect.) This was the purpose of the heater in old glass vacuum tubes. It heated the metal cathode so that it would release electrons.

$\endgroup$
  • $\begingroup$ Ok but if my source produces N photons the energy the electron receives is Nhf (with f the frequency of the photons - suppose a monochromatic source). So why the electron can't use all this energy to escape? $\endgroup$ – Federico Esposito Oct 29 '15 at 17:04
  • $\begingroup$ Just added that answer at the bottom. $\endgroup$ – MaxW Oct 29 '15 at 17:11
  • $\begingroup$ Rethinking about it: is not easier for the metal plate to emit photons in order to cool itself, before getting hot enough to ionize itself? Is there maybe a competition inside the metal plate between the number of incident photons per second (so the energy that heat the plate) and the emission of photons (maybe the thermal emission like the black body one?) that cools the plate? $\endgroup$ – Federico Esposito Oct 29 '15 at 17:19
  • $\begingroup$ As a material gets hot it emits electromagnetic waves over a broad energy range in a process known as black body radiation. That would cool the material. So you're right the incident photons on the material and the photons from black body radiation away from the material have exactly the opposite effect on the temperature. (Obviously if something gets hot in air then there are thermal currents which remove heat too.) $\endgroup$ – MaxW Oct 29 '15 at 17:27
0
$\begingroup$

The energy of a electromagnetic wave is completely independent of its amplitude. The amplitude merely describes how many quanta of energy are propagated. In essence the higher the intensity the more photons with the certain energy.

A good analogy for the photoelctric effect is a big room with an infinite amount of kids in it. They need to pay 5€ to get out of the room. If someone throws in 10 cent pieces no child will ever have enough money to escape the room because there are an infinite amount of children. If I throw in 5€ bills however a child can take it and instantly has enough money to leave the room.

In the analogy the value of the money is the energy and the amount of bills or coins is the intensity.

When the photoelectric effect was first examined the common understanding was much like yours that if the light behaves like a wave the metal should be able to accumulate energy over time until it emitted an electron but this was not the case because there is usually only single photon absorption by an electron. This photon therefore needs the right amount of energy.

$\endgroup$
  • $\begingroup$ I really don't get the wave representation of light: if I have a monochromatic source the amplitude of my wave will be the number of photons per second the source produces; if I try to picture this wave as a sinusoidal wave, I don't understand when the photons reach the detector what happen to the rest of the wave. I mean, a wave is a continuum of peaks and troughs, right? But the photons travel at velocity c and reach the detector all at the same moment, so is the wave just a single peak, with the amplitude as large as the number of photons the source produced in a single moment? $\endgroup$ – Federico Esposito Oct 29 '15 at 17:14
  • $\begingroup$ The photoelectric effect is always one photon in, and one photoelectron out. The only way to "add" the energy of photons is to heat the material. $\endgroup$ – MaxW Oct 29 '15 at 17:32
  • $\begingroup$ " if I try to picture this wave as a sinusoidal wave," don't think of photons like that. Imagine that each photon is a bullet. So each bullet has a certain amount of KE. // Photons are funny. Sometimes we model them as waves and sometimes as particles. To understand the photoelectric effect you need to think of the photons as particles. $\endgroup$ – MaxW Oct 29 '15 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.